Good morning, everyone.

Today we will continue with the asynchronous loop runner, which we had last week.

And I also think that we will get to the asynchronous cage runner.

That's why I brought a cage runner with me on site.

It's quite handy. And at this corner here we have milled out the ring so that you can see the rods.

So I'll show you the rows now. Then you can take a closer look and pass it on.

But I would have asked you to watch it in essence and not discuss every detail with your neighbor.

So, please.

Well, what is written today? Next, it's somehow analogous to the synchronous machine we had before.

It states the power balance and the torque.

That's a lot. And then, as with the synchronous machine, the current current of the asynchronous machine.

Yes, that's the first thing. And then comes the cage runner motor.

Good. As with the synchronous machine, we get the power by multiplying the voltage balance of the stator with the combined complex current

I1 star multiplied by three times. Because this is the equation for one phase and we have three phases.

That's why we multiply three times. And if we did that, we would have to form the real part here.

So, this is the voltage equation we had before. 431.28.

And this is the result here, multiplied by the combined complex current I1 star and the real part.

On this side, we now get three U1 times I1 star. And that is three times the real part.

R1 plus j omega l sigma 1 times I1, that's the term. Now multiplied by the combined complex current.

And then the last part, multiplied by the magnetization current, with I1 star.

If we look at this now, on this side, here from the real part, that's the same with the synchronous machine.

That's the power supply from the network. Is three times U1 times cosinus phi 1.

Phi 1 is the angle between the two voltages and current.

Here, now we come to the other terms on the other side. Here is R1 times I1 multiplied by I1 combined complex.

We already know that this multiplication, the square of the effective value, results here.

Three R1 I1 squared. If we now know that this term here is I squared and multiply omega l1 sigma with it,

then there is the j in front of it. This term is certainly not real. We can eliminate it.

Now we still have this term here. Omega l1 h times I mu times I1 combined complex.

The I mu has the two terms I1 plus I2. That means the term I1 multiplied by I1 combined complex is again I squared.

That means the term with I1 combined complex and this I1 from I mu, which is missing.

Then there is finally the second term I2 times I1 combined complex and omega l1 h combined to the impedance X1 a.

And we have solved this equation. The next step is of course to interpret the equation.

Here in front, this is the resulting power from the source, from the network. We can also quickly think about that.

We have three bars. Each bar has an omega resistance R1. R1 multiplied by I squared.

These are the losses. These are the losses of the bars. And because we have copper bars, these are the losses of copper bars.

The rest that remains here, it does not remain in the bar. It goes over the air gap to the flow.

This power is again called P delta. We called delta the size of the air gap.

And this is now the power transferred over the air gap. Is there another question?

We are now making life a little easier. But we do not lose the basic knowledge if we now

neglect the resistance of the bars R1. And this can be done especially with very large machines.

The influence of R1 is relatively small. And then, if R1 is equal to zero, then we can also write down,

then the resulting power is equal to the air gap power. And then the air gap power is 3 U1 I1 cosine phi.

Now comes the main work. We have to solve this term that we have just developed.

And we will try to replace I2 by I1. This is the flow voltage equation, equation 31.

This is the flow voltage equation. The voltage at the flow is zero. The resistance through S,

current I2, here it is, and here I mu. We have to think again that I1 plus I2 is contained.

And that is immediately added to the equation. And now we can summarize the terms.

The first term R2 bar through S times I2 bar, that is this. Then we summarize the term J omega L2 sigma times I2.

And in this term omega L1H times I2 bar. We summarize these two terms.

We summarize the inductivity L1H, the main inductivity, and the spread inductivity L2 sigma,